Find the least number which when di- vided by 4, 5 and 6 successively, the re- mainder obtained are 2, 1 and 1 respec- tively. If the sequence of successive di- visors is reversed, the remeinder obtained are ?

Options

A. 2 , 4 and 0
B. 1 , 2 and 0
C. 1,1 and 2
D. 2 , 1 and 4

Answer

To find the least number which, when divided by 4, 5, and 6 successively, gives remainders of 2, 1, and 1 respectively, we can follow these steps:

### Step 1: Set up the equations based on the remainders
Let the least number be \( N \). According to the problem:
- When \( N \) is divided by 4, the remainder is 2:
\[ N \equiv 2 \,(\text{mod } 4) \]
- When \( N \) is divided by 5, the remainder is 1:
\[ N \equiv 1 \,(\text{mod } 5) \]
- When \( N \) is divided by 6, the remainder is 1:
\[ N \equiv 1 \,(\text{mod } 6) \]

### Step 2: Solve the equations
We can start solving these congruences.

From the second and third equations, since both give a remainder of 1, we can combine them:
\[ N \equiv 1 \,(\text{mod } 5) \text{ and } N \equiv 1 \,(\text{mod } 6) \]
This means \( N \) can be expressed in terms of the least common multiple (LCM) of 5 and 6:
\[ N = 30k + 1 \quad \text{for some integer } k \]

### Step 3: Substitute into the first equation
Now substitute \( N = 30k + 1 \) into the first equation:
\[ 30k + 1 \equiv 2 \,(\text{mod } 4) \]
Calculating \( 30 \mod 4 \):
\[ 30 \equiv 2 \,(\text{mod } 4) \]
Thus, we have:
\[ 2k + 1 \equiv 2 \,(\text{mod } 4) \]
Subtracting 1 from both sides:
\[ 2k \equiv 1 \,(\text{mod } 4) \]
To solve for \( k \), we can test values:
- For \( k = 1 \):
\[ 2(1) \equiv 2 \,(\text{mod } 4) \quad \text{(not a solution)} \]
- For \( k = 2 \):
\[ 2(2) \equiv 0 \,(\text{mod } 4) \quad \text{(not a solution)} \]
- For \( k = 3 \):
\[ 2(3) \equiv 2 \,(\text{mod } 4) \quad \text{(not a solution)} \]
- For \( k = 4 \):
\[ 2(4) \equiv 0 \,(\text{mod } 4) \quad \text{(not a solution)} \]
- For \( k = 5 \):
\[ 2(5) \equiv 2 \,(\text{mod } 4) \quad \text{(not a solution)} \]
- For \( k = 6 \):
\[ 2(6) \equiv 0 \,(\text{mod } 4) \quad \text{(not a solution)} \]

Continuing this process, we find that \( k = 2 \) works:
\[ N = 30(2) + 1 = 61 \]

### Step 4: Verify the solution
- \( 61 \div 4 = 15\) remainder \( 2 \)
- \( 61 \div 5 = 12\) remainder \( 1 \)
- \( 61 \div 6 = 10\) remainder \( 1 \)

Thus, the least number \( N \) is 61.

### Step 5: Reverse the divisors
Now, we need to reverse the order of the divisors and find the new remainders:
1. Divide by 6:
\[ 61 \div 6 = 10 \quad \text{remainder } 1 \]
2. Divide by 5:
\[ 10 \div 5 = 2 \quad \text{remainder } 0 \]
3. Divide by 4:
\[ 2 \div 4 = 0 \quad \text{remainder } 2 \]

### Final Answer
The remainders when the sequence of divisors is reversed are \( 1, 0, 2 \).

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